What volume of 3.00M HCl in liters is needed to react completely (with nothing left over) with 0.750L of 0.400M Na2CO3?
Na2CO3 + 2 HCl --> CO2 + H2O + 2 NaCl
How many mols of Na2CO3 do we have?
.75* .4 = .3 mols
so we need .6 mols of HCl
We have 3 mols in every liter
.6 mols/ 3 mols/liter = .2 liters
To determine the volume of 3.00M HCl required to completely react with 0.750L of 0.400M Na2CO3, we can use the balanced chemical equation:
2 HCl + Na2CO3 -> 2 NaCl + H2O + CO2
From the equation, we can see that 2 moles of HCl react with 1 mole of Na2CO3.
First, let's calculate the number of moles of Na2CO3 in 0.750L of 0.400M solution:
moles of Na2CO3 = volume (L) x molarity (mol/L)
= 0.750L x 0.400mol/L
= 0.300 mol
Since 2 moles of HCl react with 1 mole of Na2CO3, we need twice the number of moles of HCl:
moles of HCl needed = 2 x moles of Na2CO3
= 2 x 0.300 mol
= 0.600 mol
Now, we can calculate the volume of 3.00M HCl needed using the molarity and the number of moles:
volume (L) = moles / molarity
= 0.600 mol / 3.00 mol/L
= 0.200 L or 200 mL
Therefore, 0.200 liters (200 mL) of 3.00M HCl is needed to react completely with 0.750L of 0.400M Na2CO3.
To find the answer, we need to use the concept of stoichiometry and the balanced chemical equation that describes the reaction between HCl and Na2CO3.
The balanced chemical equation is:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
From the balanced equation, we can see that 2 moles of HCl are required to react with 1 mole of Na2CO3.
Step 1: Calculate the number of moles of Na2CO3:
moles of Na2CO3 = volume (in liters) × molarity
moles of Na2CO3 = 0.750 L × 0.400 mol/L
moles of Na2CO3 = 0.300 moles
Step 2: Use the stoichiometry of the balanced equation to determine the number of moles of HCl required:
moles of HCl = 2 × moles of Na2CO3
moles of HCl = 2 × 0.300 moles
moles of HCl = 0.600 moles
Step 3: Calculate the volume of 3.00M HCl required:
volume (in liters) = moles of HCl / molarity
volume (in liters) = 0.600 moles / 3.00 mol/L
volume (in liters) = 0.200 L
Therefore, 0.200 liters (or 200 mL) of 3.00M HCl is needed to completely react with 0.750L of 0.400M Na2CO3.